Question Number 100557 by Mikael_786 last updated on 27/Jun/20

Ω=∫_0 ^∞  (e^(ax) /(e^(bx) +1))dx, b>a

Answered by mathmax by abdo last updated on 27/Jun/20

Ω =∫_0 ^∞  (e^(ax) /(e^(bx)  +1))dx⇒Ω =∫_0 ^∞  (e^((a−b)x) /(1+e^(−bx) ))dx =∫_0 ^∞  e^(−(b−a)x)  Σ_(n=0) ^∞  (−1)^n  e^(−nbx)  dx  =Σ_(n=0) ^∞  (−1)^n  ∫_0 ^∞  e^(−(b−a+nb)x)  dx =Σ_(n=0) ^∞  (−1)^n  [((−1)/(b−a+nb)) e^(−(b−a+nb)x) ]_0 ^∞   =Σ_(n=0) ^∞  (((−1)^n )/(b−a +nb))  ....be continued....

Answered by mathmax by abdo last updated on 27/Jun/20

let try another way   we do the changement  e^(bx)  =t ⇒bx =lnt ⇒x =((lnt)/b) ⇒  Ω =∫_0 ^∞   (e^((a/b)lnt) /(t+1)) ×(dt/(bt)) =(1/b)∫_0 ^∞    (t^((a/b)−1) /(t+1))dt    for  0<(a/b)<1  we get    Ω =(1/b)×(π/(sin(((πa)/b))))    ( i have used ∫_0 ^∞  (t^(α−1) /(1+t))dt =(π/(sin(πα)))) ⇒  Ω =(π/(b sin(((πa)/b))))

Commented byMikael_786 last updated on 28/Jun/20

thank you